Let $R$ be the region enclosed by $y=3\sqrt{4-x}$ and the axes in the first quadrant. $y$ $x$ $(0,6)$ $(4,0)$ $ R$ ${y=3\sqrt{4-x}}$ Region $R$ is the base of a solid. For each $y$ -value, the cross section of the solid taken perpendicular to the $y$ -axis is a rectangle whose base lies in $R$ and whose height is $y$. Which one of the definite integrals gives the volume of the solid? Choose 1 answer: Choose 1 answer: (Choice A) A $\int_0^4 3\sqrt{4-y}\cdot y\,dy$ (Choice B) B $\int_0^6 3\sqrt{4-y}\cdot y\,dy$ (Choice C) C $\int_0^6 \left(4-\dfrac{y^2}{9}\right)\cdot y\,dy$ (Choice D) D $\int_0^4 \left(4-\dfrac{y^2}{9}\right)\cdot y\,dy$
Explanation: Let's imagine the solid is made out of many thin slices that are perpendicular to the $y$ -axis. $y$ $x$ $(0,6)$ $(4,0)$ Each slice is a prism. Let the width of each slice be $dy$ and let the area of the prism's face, as a function of $y$, be $A(y)$. Then, the volume of each slice is $A(y)\,dy$, and we can sum the volumes of infinitely many such slices with an infinitely small width using a definite integral: $\int_a^b A(y)\,dy$ What we now need is to figure out the expression of $A(y)$ and the interval of integration. Let's consider one such slice. $y$ $x$ $(0,6)$ $(4,0)$ ${y=3\sqrt{4-x}}$ $ b(y)$ $ h(y)$ $ dy$ $ A(y)$ The face of that slice is a rectangle with base $b(y)$ and height $h(y)$. We are given that $h(y)=y$. The base $b(y)$ is equal to the distance from the curve $y=3\sqrt{4-x}$ to the $y$ -axis. To find an expression for $b(y)$, we must rewrite $y=3\sqrt{4-x}$ as a function of $y$ (which means we need to isolate $x$ ). $x=4-\dfrac{y^2}{9}$ So, for each $y$ -value, this is the base of the rectangle: $b(y)=4-\dfrac{y^2}{9}$ Now we can find the area of the rectangle: $\begin{aligned} &\phantom{=}A(y) \\\\ &=b(y)h(y) \\\\ &=\left(4-\dfrac{y^2}{9}\right)\cdot y \end{aligned}$ The bottom endpoint of $R$ is at $y=0$ and the top endpoint is at $y=6$. So the interval of integration is $[0,6]$. Now we can express the definite integral in its entirety! $\int_0^6 \left(4-\dfrac{y^2}{9}\right)\cdot y\,dy$